coder_in_use's blog

Functions with interesting properties.
By coder_in_use, history, 6 years ago, In English

I'm trying to find the function with some properties:
1. f(u) = 0 iff v = 0
2. f(av) = |a| f(v)
Does these kind of functions exist and if yes, how do they called and where can I get more info about it?

Thanks in advance!

  • Vote: I like it
  • 0
  • Vote: I do not like it

Comments Comments (5)
| Write comment?
»
6 years ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

Yes. f(x) = |x|
And i think that first property is redundantly since f(0) = f(0 * x) = |0| * f(x) = 0

→ Reply
»
6 years ago, # |
  Vote: I like it -9 Vote: I do not like it

yeah

f(x) = 0

→ Reply
»
6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The "iff" in the first property means "if and only if". This is an equivalence property which can be rewritten as

  1. If u = 0, then f(u) = 0.

  2. If f(u) = 0, then u = 0.

The contrapositives of these two conditional statements are:

  1. If f(u)  ≠  0, then u  ≠  0.

  2. If u  ≠  0, then f(u)  ≠  0.

→ Reply
»
6 years ago, # |
← Rev. 3   Vote: I like it +22 Vote: I do not like it

Here is short solution to your functional equation:

Let f(1) = c for some constant c ≠ 0. Then using (2), f(a·1) = |a|·f(1) = c|a|. Therefore, functions which satisfy the properties must be of the form f(x) = c|x| .

Checking these solutions, it is clear that f(0) = c|0| = 0 and since c ≠ 0, (1) is satisfied. Additionally, f(av) = c|av| = c|a||v| = |a|f(v) by property of absolute values.

Therefore, all work.

Notably, f(x) = 0 (as proposed by farmersrice) does not work as it fails the "only if" portion of (1).

Additionally, the first condition is not redundant for the same reason: without it, we may only conclude the "if" and not the "only if" from (2).

→ Reply